Question

a 20 liter closed vessel contains a mixture of three gasses: 2.0 mol% of CO2, 20mole%O2, and 78 mol% Nitrogen at 20 degree.

1）If the total pressures in the container is 760mmHg and the water vapor pressure at that temperature of 30.2 mmHg, calculate the partial pressure of oxygen gas in this container

2）calculate the number of grams of O2 gas in this container

Answer 1

Solution :-

1.a) Given data -----> P_T = 760 mm of Hg   

P_H2O = 30.2 mm of Hg   

Our first step is to calculate the Partial pressure of O₂

     Total pressure in the container = Partial pressure of H₂O + Partial pressure of O₂

P_T   = P_H2O  + P_O2

substitute data into formula

P_O2 = 760 - 30.2

partial pressure of O₂ = 729.8 mm of Hg

2. a)

given data :-

partial pressure of O₂ = 729.8 mm of Hg =0.960 atm

   v= 20 L

T =20+273=293 K

   n= 0.2 moles

   R= 0.0821 L-atm/mol.K

Ideal gas formula PV = nRT

p= pressure v= volume R= universal gas constant T= temprature

n (moles) = weight(W) / (M)molecular weight (32 g/mol)

   n= ( PV / RT ) = ( 0.960 x 20) / ( 0.0821 x 293 )

W = 0.7982 x M

= 0.7982 x 32

      W  = 25.5 g